Plzzz help.
the following figure shows a long wire carrying a current of 30A. The rectangular loop carries a current of 20A. Calculate the resultant force acting on the loop. Assume that a=1cm, b=8cm and l=30cm.

- ajprincess

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- ajprincess

|dw:1337319857835:dw|

- anonymous

What is B?

- ajprincess

u mean my answer for B?

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## More answers

- anonymous

I think it will involve integration.

- ajprincess

|dw:1337320632924:dw|
This is to find the force on the second current conductor.

- ajprincess

Sorry so bad at drawing. :(

- anonymous

ah ok i think i found it in book
i wil lattach 2 files hold on

- anonymous

##### 1 Attachment

- anonymous

##### 1 Attachment

- anonymous

hope it helps

- ajprincess

Ya. it helps a lot. Bt y hav they given the value of l. It is askd to find the resultant force.

- anonymous

i is amps. Is that what you mean?

- ajprincess

No. In ur attachment the question is to fin the magnetic force on the top of the wire b. And in the equation given in the solution there is no l bt only a nd b r there. in my question the value of l is also given. I am wondering y that was given and the magnetic force askd to b found here is the resultant.

- ajprincess

i mean top of the wire of length b.

- anonymous

Not sure just solve for whatever variable or convert. :)
but the diagram is almost exactly the same :)

- ajprincess

ha k. thannnnxxxxxxxx a lot.

- anonymous

:)

- ujjwal

Is your answer equal to 0.0032 N ?

- ajprincess

i get a different answer. Bt can u pls show me hw u got ur answer if u dnt mind?

- ajprincess

@ujjwal

- ujjwal

sure, the |dw:1337350437168:dw| there exists an attractive force F1 and a repulsive force F2 between the wires carrying 30 amp and 20 amp.. Calculate both forces and find their difference.. that's all..

- ajprincess

which formula did u use here to find it?

- ajprincess

did u round it off?

- ajprincess

@ujjwal

- ajprincess

i got the answer as 0.00315N.

- ujjwal

It involves large numbers.. your answer seems to be correct.. I used the relation\[F=mu _{0}I _{1} I _{2}l/2pi r \]

- ujjwal

what's wrong with OS? but i guess you got the relation i used...

- ajprincess

ya. do u knw how to derive this relation from this
|dw:1337397048799:dw|

- ajprincess

|dw:1337397220582:dw|

- ujjwal

I guess i can't see the whole relation.. I can't see it at rightmost part.. Is there a lower limit?? It's a line integral in closed path , isn't it? why do you need limits?? I can't see the whole relation though..

- ajprincess

|dw:1337397679038:dw|

- ajprincess

|dw:1337397784654:dw|
This formula is to find the force on the second conductor. Can u @Shivam_bhalla explain hw it becomes |dw:1337397890531:dw|. Plzzzz.

- anonymous

@ajprincess , Take a look here -->http://cnx.org/content/m31103/latest/

- anonymous

The above link for magnetic field due to a wire and its derivation. Now I will explain that concept which you want me to do so.

- anonymous

|dw:1337353084406:dw|
Consider 2 parallel wires carrying current I_1 and I_2
Force exerted by wire 1 on wire 2 is
F= (B_1) (I_2)(distance between two wires)

- anonymous

Sorry I wrote that in a hurry. It should be
F= (B_1) (I_2)(length of wire 2)
Now as in the previous link I gave to you, we see that
\[B_1 = \frac { \mu_0 I_1 }{2 \pi r}\]
\[F=B_1I_2(length \space of \space wire \space 2)\]
\[\frac{F}{length \space of \space wire \space 2} = \frac { \mu_0 I_1 }{2 \pi r} I_2\]
\[f=\frac { \mu_0 I_1I_2}{2 \pi r} \]
where\[ f = \frac{F}{length \space of \space wire \space 2}\]

- ajprincess

Thanxxxx a lot @Shivam_bhalla

- ajprincess

Thanxx a lot @ujjwal

- anonymous

yw :)

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